The sum of all three digited numbers that can be formed from the digits 1  to 9 and when the middle digit is perfect square is

The units place can take any of the 9 digits
The hundreds place can take any of the 9 digits
The tens place can take 3 digits (1, 4 and 9)
Total number of these numbers is =9$\times$9$\times$3 =243
When we fix the units digit the total number of possible permutations of the tens and hundreds digits is 27
Therefore each units digit occurs 27 times
Sum of all units digits is therefore 27×(1+2+...+9)
Similarly, sum of all tens digits is therefore 81×(1+4+9)
And sum of all hundreds digits is therefore 27×(1+2+....+9)
The sum of all the numbers is thus 
27×(1+2...9)+10×(81×(1+4+9))+100×(27×(1+2+....+9))=134055